Figure 24-68 shows a hemisphere with a charge of 4.00μC distributed uniformly through its volume. The hemisphere lies on an xy plane the way half a grapefruit might lie face down on a kitchen table. Point P is located on the plane, along a radial line from the hemisphere’s center of curvature, at radial distance15cm .What is the electric potential at point P due to the hemisphere?

Short Answer

Expert verified

The electric potential at point P isV=0.24MV

Step by step solution

01

Step 1: Given data:

The charge on the hemisphere,q=4.0μC=4.0×106C.

The radius of the curvature, R=15cm=0.15m.

02

Determining the concept:

The electric potential at point P will be the same as on its surface because the charges are uniformly distributed.

Formula:

The potential is define by,

V=14πεoqR=kqR

Here, ε0 is the permittivity of free space, q is the electron charge, and R is the distance, and k is the Coulomb’s constant having a value 9×109Nm2/C2.

03

Determining the electric potential:

Write the equation for the potential as below.

V=kqR

Substitute known values in the above equation.

V=(9×109Nm2/C2)×4.0×106C0.15m=2.4×105V=0.24MV

Hence, the electric potential is 0.24MV.

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